When a Homotopy Moves the Base Point
A homotopy between maps need not keep a chosen point fixed. If \(f,g:X\to Y\) are joined by a homotopy, the image of \(x_0\in X\) may travel along a path in \(Y\). Consequently, the loops obtained by applying \(f\) and \(g\) to a loop based at \(x_0\) are based at possibly different points. They belong to different fundamental groups, so comparing their classes requires a path that transports one base point to the other.
We use the convention that \(\alpha*\beta\) means first traverse \(\alpha\), then traverse \(\beta\). The reverse of a path \(\gamma:I\to Y\) is \(\bar\gamma(s)=\gamma(1-s)\). A loop at \(y\) is a path that begins and ends at \(y\), and \(\pi_1(Y,y)\) is the group of based homotopy classes of loops at \(y\).
For a based map \(f:(X,x_0)\to(Y,y_0)\), the induced homomorphism \(f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)\) sends \([\alpha]\) to \([f\circ\alpha]\). When the base point changes along a path \(h\) from \(y_0\) to \(y_1\), the base-point change map is \(\beta_h:\pi_1(Y,y_0)\to\pi_1(Y,y_1)\), given by \(\beta_h([\alpha])=[\bar h*\alpha*h]\). The path starts at \(y_1\), reaches \(y_0\), traverses \(\alpha\), and returns to \(y_1\). This map is an isomorphism; its inverse is \(\beta_{\bar h}\), as the adjacent path and reverse-path segments contract by retracing. The base-point change results establish that this construction respects based homotopy classes.
The Track Formula
In concrete terms, for every loop \(\alpha\) at \(x_0\), the loop \(\bar h*(f\circ\alpha)*h\), based at \(g(x_0)\), is based-homotopic to \(g\circ\alpha\). The proof builds this homotopy from the square traced by \(\alpha\) during \(H\); it does not assume that the base point stays fixed.
Proof. Fix a loop \(\alpha:I\to X\) at \(x_0\). For each \(t\in I\), define \(\alpha_t(s)=H(\alpha(s),t)\). This is a path from \(h(t)\) to \(h(t)\); in particular, \(\alpha_0=f\circ\alpha\) and \(\alpha_1=g\circ\alpha\). Also define the remaining part of the track by \(h_t(v)=h(t+(1-t)v)\). It runs from \(h(t)\) to \(g(x_0)\). Its reverse runs from \(g(x_0)\) to \(h(t)\).
Construct a loop \(\lambda_t\), always based at \(g(x_0)\), by traversing \(\bar h_t\), then \(\alpha_t\), then \(h_t\). To see that these loops form a based homotopy, parameterize each of the three paths on one third of \(I\): \[ \lambda_t(u)= \begin{cases} h\bigl(t+(1-t)(1-3u)\bigr),&0\leq u\leq 1/3,\\ H\bigl(\alpha(3u-1),t\bigr),&1/3\leq u\leq 2/3,\\ h\bigl(t+(1-t)(3u-2)\bigr),&2/3\leq u\leq 1. \end{cases} \] At \(u=1/3\), the first and second expressions both equal \(h(t)\), because \(\alpha(0)=x_0\). At \(u=2/3\), the second and third expressions both equal \(h(t)\), because \(\alpha(1)=x_0\). Each piece is continuous, and the pieces agree where they meet, so \((u,t)\mapsto\lambda_t(u)\) is continuous. Each \(\lambda_t\) begins and ends at \(g(x_0)\).
At \(t=0\), the three pieces traverse \(\bar h\), \(f\circ\alpha\), and \(h\), so \(\lambda_0=\bar h*(f\circ\alpha)*h\) with the stated parameterization. At \(t=1\), the first and third pieces are constant at \(g(x_0)\), and the middle piece traverses \(g\circ\alpha\). Thus \(\lambda_1\) is \(g\circ\alpha\) with constant pauses at its endpoints. Removing those pauses is a based homotopy: if \(\rho(u)=0\) for \(u\leq1/3\), \(\rho(u)=3u-1\) for \(1/3\leq u\leq2/3\), and \(\rho(u)=1\) for \(u\geq2/3\), then \(\lambda_1(u)=(g\circ\alpha)(\rho(u))\). The maps \((g\circ\alpha)((1-r)\rho(u)+ru)\), for \(r\in I\), give a based homotopy from \(\lambda_1\) to \(g\circ\alpha\). Therefore \([\bar h*(f\circ\alpha)*h]=[g\circ\alpha]\). This holds for every loop \(\alpha\), proving the equality of homomorphisms. \(\square\)
Worked Examples
Worked Example: A Rotation of the Circle
Give \(S^1=\{z\in\mathbb C:|z|=1\}\) its subspace topology from \(\mathbb C\). Define \(H:S^1\times I\to S^1\) by \(H(z,t)=e^{i\pi t}z\). Take \(x_0=1\). Find the maps at the two ends and the base-point track.
At \(t=0\), \(H(z,0)=e^0z=z\), so \(f(z)=z\). At \(t=1\), \(H(z,1)=e^{i\pi}z=-z\), so \(g(z)=-z\). The track is \(h(t)=H(1,t)=e^{i\pi t}\). Its endpoints are \(h(0)=1=f(1)\) and \(h(1)=-1=g(1)\).
For the loop \(\alpha(s)=e^{2\pi i s}\) at \(1\), the two image loops are \((f\circ\alpha)(s)=e^{2\pi i s}\), based at \(1\), and \((g\circ\alpha)(s)=-e^{2\pi i s}=e^{i\pi}e^{2\pi i s}\), based at \(-1\). The transported loop is \(\bar h*(f\circ\alpha)*h\): it starts at \(-1\), follows \(\bar h(s)=e^{i\pi(1-s)}\) to \(1\), traverses \(e^{2\pi i s}\), and follows \(h(s)=e^{i\pi s}\) back to \(-1\). The track formula gives \([\bar h*(f\circ\alpha)*h]=[g\circ\alpha]\) in \(\pi_1(S^1,-1)\). In particular, the endpoints and base point of every piece match the required concatenation order.
Check: \(e^{i\pi}=\cos\pi+i\sin\pi=-1\), so the endpoint of the track is indeed \(-1\), and \(g(\alpha(0))=g(1)=-1\). Both sides of the asserted equality are therefore classes of loops based at \(-1\).
Worked Example: Moving Between Two Points in the Plane
Give \(\mathbb R^2\) its standard topology. Let \(X=S^1\), and set \(H(z,t)=(2t,0)\) for every \(z\in S^1\). This is a homotopy from the constant map \(f(z)=(0,0)\) to the constant map \(g(z)=(2,0)\). For any \(x_0\in S^1\), the track is \(h(t)=(2t,0)\).
For any loop \(\alpha\) at \(x_0\), \(f\circ\alpha\) is the constant loop at \((0,0)\), while \(g\circ\alpha\) is the constant loop at \((2,0)\). The transported loop \(\bar h*(f\circ\alpha)*h\) starts at \((2,0)\), travels along the line segment to \((0,0)\), stays there during the constant-loop part, and retraces the segment to \((2,0)\).
To check its class directly, write this transported loop as \(\gamma:I\to\mathbb R^2\). The formula \(C(r,s)=(2,0)+(1-r)(\gamma(s)-(2,0))\), for \(r,s\in I\), is a continuous based homotopy: \(C(0,s)=\gamma(s)\), \(C(1,s)=(2,0)\), and \(C(r,0)=C(r,1)=(2,0)\). Thus the transported loop is based-homotopic to the constant loop \(g\circ\alpha\), as the theorem predicts. The example also checks the degenerate case in which both induced maps have only constant loops in their images.
Worked Example: A Fixed Base Point Within a Changing Map
Again give \(S^1\) its subspace topology from \(\mathbb C\). Define \(H(z,t)=e^{i\pi t(1-\operatorname{Re}z)}z\), and choose \(x_0=1\). The formula is continuous because real part, multiplication, and the exponential are continuous. At \(t=0\), \(f(z)=z\); at \(t=1\), \(g(z)=e^{i\pi(1-\operatorname{Re}z)}z\).
At the chosen point, \(\operatorname{Re}(1)=1\), so \(h(t)=H(1,t)=e^{i\pi t(1-1)}\cdot1=e^0=1\) for every \(t\). Thus \(f(1)=g(1)=1\), and the track is constant. Base-point change along a constant path is the identity on loop classes: adding constant pauses to a loop does not change its based homotopy class. The track formula therefore gives \(f_*=g_*\) as homomorphisms from \(\pi_1(S^1,1)\) to itself, even though the endpoint maps need not be the same at other points.
Check: substituting \(z=1\) into both endpoint maps gives \(f(1)=1\) and \(g(1)=e^{i\pi(1-1)}\cdot1=1\). Hence the two induced maps really do have the same target fundamental group.
A Common Base-Point Mistake
A frequent error is to write \(f_*=g_*\) as soon as \(f\) and \(g\) are homotopic. If \(f(x_0)\ne g(x_0)\), then \(f_*\) takes values in \(\pi_1(Y,f(x_0))\), whereas \(g_*\) takes values in \(\pi_1(Y,g(x_0))\). These are not even maps with the same codomain. In the circle rotation example, \(f_*\) has codomain \(\pi_1(S^1,1)\) and \(g_*\) has codomain \(\pi_1(S^1,-1)\). The correction is to transport the first output along the actual track: \(\beta_h\circ f_*=g_*\).
The direction of the track matters too. If \(h\) runs from \(y_0\) to \(y_1\), then \(\bar h*\alpha*h\) is based at \(y_1\), not \(y_0\): its first segment starts at \(y_1\), and its last segment ends there. Reversing the order would fail to join paths at their endpoints. The square proof explains why this particular transport works: each intermediate loop uses the part of the track from its current base point to the final one.
This distinction is useful whenever maps are compared through a homotopy but the image of a chosen point is not fixed. It separates two questions: what the homotopy does to loops, and how loop classes are identified when their base points differ. When the track is constant, the formula reduces to the based-homotopy result; when it moves, base-point change supplies the precise comparison.
Check Your Understanding
For each question, keep track of the starting point and ending point of every path.
- If \(h\) runs from \(y_0\) to \(y_1\), at which point is the loop \(\bar h*\alpha*h\) based, when \(\alpha\) is based at \(y_0\)?
- For a homotopy \(H\) and a loop \(\alpha\) at \(x_0\), write the loop at \(g(x_0)\) whose class is compared with \([g\circ\alpha]\).
- In the circle rotation example, compute \(h(0)\), \(h(1)\), \(f(1)\), and \(g(1)\).
- Why is the assertion \(f_*=g_*\) not properly typed when \(f(x_0)\ne g(x_0)\)?
- What does the track formula become when \(H(x_0,t)\) is constant in \(t\)?